CodeForces 560A

题意：给定一个货币系统，问不能组成的最小的钱数是多少。

思路：水，只要检查有没有出现1即可，有则输出-1，否则1.

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000+5;
const int M=1e3+5;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int v[M],vis[N];
int main()
{int n;cls(vis,0);scanf("%d",&n);ft(i,1,n) scanf("%d",&v[i]),vis[v[i]]=1;sort(v+1,v+1+n);if (v[1]==1) puts("-1");else puts("1");
}

CodeForces 560B

题意：给定一个相框，问能否把两张照片放进去。。

思路：依旧水，对于两个照片，只有长长，短短，长短和短长这四种放法，没swap会wa？？

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int a[5],b[5];
bool ok(){int s1,s2;s1=max(b[3],b[2]);s2=a[3]+a[2];if (s1>s2) swap(s1,s2);if (s1<=a[1]&&s2<=b[1]) return 1;s1=max(a[3],a[2]);s2=b[3]+b[2];if (s1>s2) swap(s1,s2);if (s1<=a[1]&&s2<=b[1]) return 1;s1=max(b[3],a[2]);s2=a[3]+b[2];if (s1>s2) swap(s1,s2);if (s1<=a[1]&&s2<=b[1]) return 1;s1=max(a[3],b[2]);s2=b[3]+a[2];if (s1>s2) swap(s1,s2);if (s1<=a[1]&&s2<=b[1]) return 1;return 0;
}
int main()
{ft(i,1,3) {scanf("%d %d",&a[i],&b[i]);if (a[i]>b[i]) swap(a[i],b[i]);}if(ok()) puts("YES");else puts("NO");
}

CodeForces 560C

题意：给定一个六边形的各边长度，已知相邻两边夹角均为120度，求能分成多少变长为1的正三角形。

思路：分别延长v2，v4，v6得到一个大的正三角形，用两个边的平方表示正三角形面积（多乘了2/根号3），然后减去多补的那部分。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000+5;
const int M=1e5+5;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define ft(i,s,n) for (int i=s;i<=n;i++)
#define frt(i,s,n) for (int i=s;i>=n;i--)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int v[8];int main()
{ft(i,1,6) scanf("%d",&v[i]);int s=(v[1]+v[2]+v[3])*(v[1]+v[2]+v[3])-v[1]*v[1]-v[3]*v[3]-v[5]*v[5];printf("%d\n",s);
}